A bisector ЕF is drawn in a right-angled triangle DCE with a right angle C, and FC = 13 cm

A bisector ЕF is drawn in a right-angled triangle DCE with a right angle C, and FC = 13 cm Find the distance from point F to line DE.

Let us prove that the triangles EFC and EFH are equal.

Both triangles are rectangular. Let the angle A = X0, then the angle ACB is equal to (90 – X).

Since, by condition, EF is the bisector of angle E, then the angle HEF = CEF. The hypotenuse EF is common for both triangles, therefore, the triangles EFC and EFH are equal in hypotenuse and acute angle.

Then the leg FC = FH = 13 cm, since they lie opposite the same angles.

Answer: The distance from point F to line DE is 13 cm.